Find the Inverse y=-4x^2-2

$y = - 4 x^{2} - 2$

Interchange the variables.

$x = - 4 y^{2} - 2$

Solve for $y$ .

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Rewrite the equation as $- 4 y^{2} - 2 = x$ .

$- 4 y^{2} - 2 = x$

Add $2$ to both sides of the equation.

$- 4 y^{2} = x + 2$

Divide each term by $- 4$ and simplify.

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Divide each term in $- 4 y^{2} = x + 2$ by $- 4$ .

$\frac{- 4 y^{2}}{- 4} = \frac{x}{- 4} + \frac{2}{- 4}$

Cancel the common factor of $- 4$ .

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Cancel the common factor.

$\frac{- 4 y^{2}}{- 4} = \frac{x}{- 4} + \frac{2}{- 4}$

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{x}{- 4} + \frac{2}{- 4}$

Simplify each term.

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Move the negative in front of the fraction.

$y^{2} = - \frac{x}{4} + \frac{2}{- 4}$

Cancel the common factor of $2$ and $- 4$ .

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Factor $2$ out of $2$ .

$y^{2} = - \frac{x}{4} + \frac{2 (1)}{- 4}$

Cancel the common factors.

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Factor $2$ out of $- 4$ .

$y^{2} = - \frac{x}{4} + \frac{2 \cdot 1}{2 \cdot - 2}$

Cancel the common factor.

$y^{2} = - \frac{x}{4} + \frac{2 \cdot 1}{2 \cdot - 2}$

Rewrite the expression.

$y^{2} = - \frac{x}{4} + \frac{1}{- 2}$

Move the negative in front of the fraction.

$y^{2} = - \frac{x}{4} - \frac{1}{2}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- \frac{x}{4} - \frac{1}{2}}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Factor $- 1$ out of $- \frac{x}{4} - \frac{1}{2}$ .

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Factor $- 1$ out of $- \frac{x}{4}$ .

$y = \pm \sqrt{- (\frac{x}{4}) - \frac{1}{2}}$

Factor $- 1$ out of $- \frac{1}{2}$ .

$y = \pm \sqrt{- (\frac{x}{4}) - (\frac{1}{2})}$

Factor $- 1$ out of $- (\frac{x}{4}) - (\frac{1}{2})$ .

$y = \pm \sqrt{- (\frac{x}{4} + \frac{1}{2})}$

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \pm \sqrt{- (\frac{x}{4} + \frac{1}{2} \cdot \frac{2}{2})}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{1}{2}$ and $\frac{2}{2}$ .

$y = \pm \sqrt{- (\frac{x}{4} + \frac{2}{2 \cdot 2})}$

Multiply $2$ by $2$ .

$y = \pm \sqrt{- (\frac{x}{4} + \frac{2}{4})}$

Combine the numerators over the common denominator.

$y = \pm \sqrt{- \frac{x + 2}{4}}$

Rewrite $- \frac{x + 2}{4}$ as ${(\frac{1^{2}}{2})}^{2} \cdot (- (x + 2))$ .

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Factor the perfect power $1^{2}$ out of $x + 2$ .

$y = \pm \sqrt{- \frac{1^{2} (x + 2)}{4}}$

Factor the perfect power $2^{2}$ out of $4$ .

$y = \pm \sqrt{- \frac{1^{2} (x + 2)}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} (x + 2)}{2^{2} \cdot 1}$ .

$y = \pm \sqrt{- ({(\frac{1}{2})}^{2} (x + 2))}$

Reorder $- 1$ and ${(\frac{1}{2})}^{2}$ .

$y = \pm \sqrt{{(\frac{1}{2})}^{2} \cdot (- 1 (x + 2))}$

Rewrite $1$ as $1^{2}$ .

$y = \pm \sqrt{{(\frac{1^{2}}{2})}^{2} \cdot (- 1 (x + 2))}$

Add parentheses.

$y = \pm \sqrt{{(\frac{1^{2}}{2})}^{2} \cdot (- (x + 2))}$

Pull terms out from under the radical.

$y = \pm \frac{1^{2}}{2} \cdot \sqrt{- (x + 2)}$

One to any power is one.

$y = \pm \frac{1}{2} \cdot \sqrt{- (x + 2)}$

Combine $\frac{1}{2}$ and $\sqrt{- (x + 2)}$ .

$y = \pm \frac{\sqrt{- (x + 2)}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{- (x + 2)}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.